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3 years ago

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a 4kg metal block absorbs 5000j of energy and increases to a temperature of 22°c. the metal has a specific heat capacity of 250j

/kg°c. what was the original temperature of the block?

1 answer:

e-lub [12.9K]3 years ago

6 0

Answer:

17 °C

Explanation:

From specific Heat capacity.

Q = cm(t₂-t₁)................. Equation 1

Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.

make t₁ the subject of the equation

t₁ = t₂-(Q/cm)............... Equation 2

Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c

Substitute into equation 2

t₁ = 22-[5000/(4×250)

t₁ = 22-(5000/1000)

t₁ = 22-5

t₁ = 17 °C

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th

jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

= $130\times 6.894$

= $896.22 \ Kpa$

or,

= $896.22\times 10^3 \ Pa$

Z₂ = 10ft

= 3.05 m

$\delta$ = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒ $\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2$

On substituting the values, we get

⇒ $\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2$

⇒ $\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05$

⇒ $P_2=866330 \ P_a$

i.e.,

⇒ $=866330\times 0.000145$

⇒ $=126 \ Psi$

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3 years ago

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Likurg_2 [28]

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

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3 years ago

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ASHA 777 [7]

Answer:

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3 years ago

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Alenkinab [10]

Answer:

The reading on the ammeter A₁ , should be 2 Amp.

Explanation:

Given circuit shows all the bulbs are connected in parallel and ammeter A(T) at the source reads 6 Amp.

So, as the bulbs are in parallel connection, the current gets divided equal to each bulb.

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But it will show 6 Amp, which is three times of the required value(2 amp).

This is because there was a mistake while making the circuit connections.

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To correct this, remove the wire connecting ammeter and the second bulb.

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3 years ago